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A Sorting Lower Bound


Date: 10/01/98 at 07:28:15
From: John Smither
Subject: Lower Bound Proof

Hi!

I have the following university problem:

Show that you cannot sort 3 elements with 2 comparisons. This is an 
example of lower bound.  

Can you help?

Thanks,
John


Date: 10/01/98 at 09:08:24
From: Doctor Rob
Subject: Re: Lower Bound Proof

There are six possible orders for the three numbers. Each comparison
splits this set of orders into two sets, one of which agrees with the
results, and one of which doesn't. Two comparisons will split the set 
of orders into four sets. At least one of these sets must contain at 
least two of the orders, by the Pigeonhole Principle. Those two cannot 
be distinguished by this procedure. You fill in the details and 
reasons.

Here is another proof. If the two comparisons only involve two of the
three elements, the second one is redundant, and the size of the third
element relative to the first two will remain unknown. Thus the two
comparisons must involve all three elements. Let the common element be 
a, and the first comparison a vs. b, the second a vs. c. No matter the
results, the relative sizes of b and c will remain unknown, unless a 
is the middle element, which cannot be guaranteed ahead of time, and 
happens only part of the time at random. We cannot improve on this by 
comparing a vs. b, and then c vs. max(a,b), since if c is smaller than 
max(a,b), we will not be able to tell whether or not c is larger than 
min(a,b). Similarly, comparing c to min(a,b) won't work either. There 
is essentially nothing else to try.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Analysis

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